A) What are the partial pressures of NH3 and H2S at equilibrium, what are the values of P(NH3) and P(H2S)?
[i have a value of 0.043 for initial pressure of H2S, am i suppose to use this?]

B) What is the mole fraction of H2S in the gas mixture at equilibrium?
[since the temps are the same and they have the same mole to mole ratio is my answer .5?]

C)What is the minimum mass of NH4HS that must be added to the 5L flask when charged with the 0.3 g of H2S at 25C to achieve equilibrium?
[no clue what to do!? any help with these questions would be great! thank you in advance]

Thank you for answering me! You don't know how much it means to me. I think I'm over thinking this but I need all the help I can get.

So for A) I just divided .120/.0431 and got 2.78. It just seems too simple to me haha.
For letter C I tried finding the moles of H2S. Since its one to one ratio wouldn't the moles be the same? Then placing these two over Kp to give me the moles of that?

No, mols are not the same. They would be if you had the pure NH4HS coming to equilibrium but you don't have that. The problem has charged the vessel with 0.300 g H2S (0.0431 moles H2S); therefore, your value of 2.78 atm is the partial pressure of NH3. Knowing p NH3 you should be able to obtain n from PV = nRT and from that you can get grams NH3.

Try this.
If the system were allowed to come to equilibrium (without the added H2S), the partial pressure of each gas would be sqrt(Kp) = about 0.35 atm. Since the added 0.300 g H2S has a pressure of 0.04315 atm initially, we note that is smaller than the H2S would have been had no H2S been added at all. Therefore, I believe the reaction will shift to the right since there is insufficient H2S.
.........NH4HS ==> NH3 + H2S
initial............0..0.04315
change............+p....+p
equil..............p..0.04315+p
Solve for p.
If I didn't make an error my number is 0.3255 for p and 0.3687 for 0.04315+p.
That seems reasonable since
(0.3255*0.3687) = 0.120. Remember I have too many s.f. in my numbers; only 3 s.f. should be used.

Idk why maybe because its 2 am and i cant do things right at this time, but am also trying to solve this problem and when i get to the part to use the quadratic formula to find p i get a different number i have it set up as this:
p^2+.04315p-120

and i get two answers for p obviously, 1 is .3686 and the other one is -.3255 (which is the same as your answer except is negative) I wonder what am doing wrong >< , thx for all your help

I re-worked the problem and I get two roots for the equation; one is -0.36866 (the same as yours but mine is minus) and +0.3255 (the same as your but mine is +). If you are solving this by hand you are slipping a sign somewhere; if you are using the calculator you are using the negative of the correct formula.